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Quantitative Aptitude: Clocks & Calendars

1. Concept Definitions & Explanations

Clocks

  • Clock Face: A circle divided into 12 equal hours and 60 equal minute spaces.
  • Hour Hand Speed: Covers $360^\circ$ in 12 hours.
    • Speed $= 30^\circ/\text{hour} = 0.5^\circ/\text{minute}$.
  • Minute Hand Speed: Covers $360^\circ$ in 60 minutes.
    • Speed $= 6^\circ/\text{minute}$.
  • Relative Speed: The minute hand gains $5.5^\circ$ per minute ($6^\circ - 0.5^\circ$) over the hour hand.

Calendars

  • Odd Days: The remainder left when total number of days is divided by 7. This is used to determine weekdays.
  • Ordinary Year: 365 days (52 weeks + 1 odd day).
  • Leap Year: 366 days (52 weeks + 2 odd days).
    • A year is a leap year if it is divisible by 4, unless it is a century year (ending in 00).
    • Century years must be divisible by 400 to be leap years. E.g., 2000 is a leap year, but 1900 is not.

2. Key Formulas & Shortcuts

Clock Angles

The angle $\theta$ between the hour hand and the minute hand at $H$ hours and $M$ minutes is: $$\theta = \left| 30H - \frac{11}{2}M \right|$$ Note: If the result is greater than $180^\circ$, find the reflex angle by subtracting it from $360^\circ$.

Calendar Odd Days Reference

  • Odd Days in Centuries:

    • 100 years $= 5$ odd days
    • 200 years $= 3$ odd days
    • 300 years $= 1$ odd day
    • 400 years $= 0$ odd days
    • Multiples of 400 years (800, 1200, 1600, 2000) have 0 odd days.

  • Odd Days in Months:

    • Jan, Mar, May, Jul, Aug, Oct, Dec (31 days) $= 3$ odd days
    • Apr, Jun, Sep, Nov (30 days) $= 2$ odd days
    • Feb (28 days) $= 0$ odd days
    • Feb (29 days) $= 1$ odd day

  • Weekday Mapping Table:

Odd Days Day of Week
0 Sunday
1 Monday
2 Tuesday
3 Wednesday
4 Thursday
5 Friday
6 Saturday

3. Practice Problems

Problem 1

Find the angle between the hour hand and the minute hand of a clock at 8:20.

Problem 2

What was the day of the week on 15th August 1947?

Problem 3

If 9th June 2026 is a Tuesday, what day of the week will 9th June 2030 be?

Problem 4

At what time between 4 and 5 o'clock will the hands of a clock be together (coincide)?

Problem 5

A clock is set right at 5 a.m. The clock loses 16 minutes in 24 hours. What will be the true time when the clock indicates 10 p.m. on the 4th day?

Problem 6

If the 1st of January 2001 was a Monday, what day of the week was the 1st of January 2002?

Problem 7

How many times do the hands of a clock stand at right angles to each other in a day (24 hours)?

Problem 8

What is the number of odd days in 400 years?

Problem 9

How many times in a day (24 hours) do the hands of a clock point in opposite directions (are collinear but not coinciding)?

Problem 10

What was the day of the week on 26th January 1950?

4. Step-by-Step Solutions

Solution 1

  1. Identify parameters:
    • $H = 8$, $M = 20$.
  2. Apply the Clock Angle Formula: $$\theta = \left| 30(8) - \frac{11}{2}(20) \right|$$ $$\theta = \left| 240 - 11 \times 10 \right|$$ $$\theta = \left| 240 - 110 \right| = 130^\circ$$
  3. Answer: The angle between the hands is $130^\circ$.

Solution 2

  1. Break down the years:
    • We need to calculate odd days up to 15th August 1947.
    • Period completed: 1946 years + period from Jan 1st to Aug 15th, 1947.
  2. Calculate odd days for 1946 years:
    • Split 1946 into parts: $1600 \text{ years} + 300 \text{ years} + 46 \text{ years}$.
      • 1600 years $= 0$ odd days (multiple of 400).
      • 300 years $= 1$ odd day.
      • 46 years contains:
        • $46 \div 4 = 11$ leap years.
        • $46 - 11 = 35$ ordinary years.
        • Odd days in 46 years $= (11 \times 2) + (35 \times 1) = 22 + 35 = 57$ days.
        • Convert to weekly remainder: $57 \div 7 = 8$ weeks + $1$ odd day.
    • Total odd days for 1946 years $= 0 + 1 + 1 = 2$ odd days.
  3. Calculate odd days for Jan 1 to Aug 15, 1947:
    • 1947 is an ordinary year (Feb has 28 days).
    • Monthly odd days:
      • Jan (3) + Feb (0) + Mar (3) + Apr (2) + May (3) + Jun (2) + Jul (3) + 15 days in August.
      • Sum $= 3 + 0 + 3 + 2 + 3 + 2 + 3 + 15 = 31$ days.
      • Convert to weekly remainder: $31 \div 7 = 4$ weeks + $3$ odd days.
  4. Total cumulative odd days:
    • Total $= 2 \text{ (from years)} + 3 \text{ (from months)} = 5$ odd days.
  5. Map to Weekday:
    • 5 corresponds to Friday.
  6. Answer: 15th August 1947 was a Friday.

Solution 3

  1. Analyze year differences from 2026 to 2030:
    • 2026 (remaining) to 2030.
    • Number of years elapsed $= 4$ years.
  2. Find the number of leap years in this span:
    • The year 2028 is a leap year.
    • Leap years $= 1$, Ordinary years $= 3$.
  3. Calculate total odd days:
    • Odd days $= (1 \times 2) + (3 \times 1) = 2 + 3 = 5$ odd days.
  4. Determine the new day:
    • Add 5 days to Tuesday:
      • Tuesday + 5 days = Sunday.
  5. Answer: 9th June 2030 will be a Sunday.

Solution 4

  1. Understand hand coincidence conditions:
    • Hands coincide when the angle $\theta = 0$.
    • The time is between 4 and 5, so $H = 4$. Let the minute be $M$.
  2. Apply the Clock Angle formula:
    • $\theta = \left| 30(4) - \frac{11}{2}M \right| \implies 0 = 30(4) - \frac{11}{2}M$
    • $\frac{11}{2}M = 120 \implies M = \frac{240}{11} = 21\frac{9}{11}$ minutes.
  3. Answer: The hands will coincide at $21\frac{9}{11}$ minutes past 4.

Solution 5

  1. Calculate total elapsed time indicated by the clock:
    • Start: 5 a.m. Day 1.
    • End: 10 p.m. Day 4.
    • Day 1 (5 a.m.) to Day 4 (5 a.m.) $= 3 \text{ days} = 72 \text{ hours}$.
    • Day 4 (5 a.m.) to Day 4 (10 p.m.) $= 17 \text{ hours}$.
    • Total indicated time $= 72 + 17 = 89$ hours.
  2. Determine the relation between incorrect and correct time:
    • The clock loses 16 minutes in 24 hours.
    • Indicated time in 24 hours $= 23 \text{ hours } 44 \text{ minutes} = 23\frac{44}{60} \text{ hours} = 23\frac{11}{15} \text{ hours} = \frac{356}{15}$ hours.
    • Thus, $\frac{356}{15}$ hours of incorrect clock $= 24$ hours of correct clock.
  3. Calculate the correct time for 89 indicated hours:
    • $\text{True time} = 89 \times \frac{24}{356/15} = 89 \times \frac{24 \times 15}{356} = 89 \times \frac{360}{356}$.
    • Note that $356 = 4 \times 89$.
    • $\text{True time} = \frac{360}{4} = 90$ hours.
  4. Determine the true end time:
    • The correct elapsed time is 90 hours (which is 1 hour more than the indicated 89 hours).
    • Therefore, the true time is 1 hour ahead of 10 p.m.
    • True time $= 11$ p.m.
  5. Answer: The true time is 11 p.m..

Solution 6

  1. Analyze year differences:
    • 2001 is an ordinary year (not divisible by 4).
    • Therefore, 2001 has 365 days.
  2. Calculate odd days:
    • 365 days $= 52 \text{ weeks} + 1 \text{ odd day}$.
  3. Determine the new day:
    • Monday + 1 odd day = Tuesday.
  4. Answer: 1st of January 2002 was a Tuesday.

Solution 7

  1. Analyze clock right angle occurrences:
    • In 1 hour, the hands are at right angles twice (except between 2-3 and 8-9, where they are at right angle only once each, or rather, the transitions at 3:00 and 9:00 are shared).
    • Specifically:
      • In a 12-hour period, right angles occur 22 times.
  2. Calculate for a full day (24 hours):
    • In 24 hours, they will be at right angles $22 \times 2 = 44$ times.
  3. Answer: The hands are at right angles 44 times in a day.

Solution 8

  1. Break down 400 years:
    • 100 years $= 5$ odd days.
    • 200 years $= 3$ odd days.
    • 300 years $= 1$ odd day.
  2. Calculate for 400 years:
    • 400 years $= 4 \times (\text{odd days in 100 years}) + 1 \text{ leap day (since the 400th year is a leap year)}$.
    • Odd days $= 4 \times 5 + 1 = 21$ days.
    • $21 \div 7 = 3$ weeks with remainder $0$ odd days.
  3. Answer: The number of odd days in 400 years is 0.

Solution 9

  1. Analyze opposite direction occurrences:
    • The hands point in opposite directions ($180^\circ$ angle) once every hour, except between 5 and 7 where they align only at exactly 6:00.
    • Thus, they point in opposite directions 11 times in 12 hours.
  2. Calculate for 24 hours:
    • Total occurrences $= 11 \times 2 = 22$ times.
  3. Answer: The hands point in opposite directions 22 times in a day.

Solution 10

  1. Break down the years:
    • Up to 26th January 1950.
    • Period completed: 1949 years + 26 days of January 1950.
  2. Calculate odd days for 1949 years:
    • $1600 \text{ years} = 0$ odd days.
    • $300 \text{ years} = 1$ odd day.
    • 49 years contains:
      • $49 \div 4 = 12$ leap years.
      • $49 - 12 = 37$ ordinary years.
      • Odd days in 49 years $= (12 \times 2) + (37 \times 1) = 24 + 37 = 61$ days.
      • $61 \div 7 = 8$ weeks + $5$ odd days.
    • Total odd days for 1949 years $= 0 + 1 + 5 = 6$ odd days.
  3. Calculate odd days for January 1950:
    • 26 days in January $= 26 \div 7 = 3$ weeks + $5$ odd days.
  4. Total cumulative odd days:
    • Total $= 6 + 5 = 11 \implies 11 \div 7 = 1$ week + $4$ odd days.
  5. Map to Weekday:
    • 4 corresponds to Thursday.
  6. Answer: 26th January 1950 was a Thursday.